Electricity is a form of energy that revolutionized the human lifestyle, today life is totally dependent on electric appliances. Electricity even has great advantages as its efficiency, easily transmitted, cleanness. Use of electricity is growing very fast with substitute of natural resources. In this chapter we shall study the basic concept of electricity, electric charge and its properties, conductor and insulator, electric potential, ohm's law, resistance and heating effect of current and its applications.
An electric circuit is a closed path of conducting wires through which electrons flow in electric devices, switching devices, source of electricity, etc.
The SI unit of electric current is Ampere. 1 ampere is defined as the flow of 1 Coulomb of charge through a wire in 1 second. Smaller unit of current is milliAmpere that is 1/1000 ampere.
Charge is flow of number of electrons.
Q=ne
Q is charge, n = number, e = electron
Whereas, One electron possesses a charge of 1.6 × 10−19 C,
i.e., 1.6 × 10−19 C of charge is contained in 1 electron.
∴ 1 C of charge is contained in
Hence, 6x1018 electrons constitute one coulomb of charge.
Potential difference across a conductor is created by devices such as cell, battery, power supply, generator etc.
Potential difference is defined as the work done to move per unit charge between the two ends of the current carrying conductor. Potential difference is designated by V.
V=W/Q
If 1 J of work is required to move a charge of amount 1 C from one point to another, then it is said that the potential difference between the two points is 1 V.
The amount of work done is equal to energy consumed. The energy given to each coulomb of charge is equal to the amount of work required to move unit charge
Potential difference =
Work Done = Potential Difference x Charge
Given,
Charge = 1 C
Potential difference = 6 V
Work Done = 6 x 1 = 6 J
Total 6 J of energy is given to each coulomb of charge passing through a battery of 6 V.
Resistance is the tendency of a resistor to oppose the flow of current.
R= ρl/A
R is resistance, ρ is resistivity, l is length of conductor and A is area of cross section of current carrying conductor.
It depends upon the following factors:
(a) Temperature of the conductor
(b) Cross-sectional area of the conductor
(c) Length of the conductor
(d) Material of the conductor
Resistance of a conductor is inversely proportional to the area of cross-section of the wire.Thicker the wire, lower is the resistance of the wire and vice-versa as current can flow more easily through a thick wire than a thin wire.
Resistance is tendency that oppose the flow of current which depends upon
R= 
l = Length of the wire
A = Area of cross-section of the wire
According to ohm's law we know that current is directly potential to the potential difference across the two ends.
V = IR
Here, R= Resistance of the electrical component
V= Potential difference and I=Current
The potential difference is reduced by half, keeping resistance constant.
Let the new resistance be R' and the new amount of current be I '.
Therefore, from Ohm’s law, we obtain the amount of new current.
So, the amount of current flowing through the electrical component is reduced by half.
An alloy has higher resistivity than the pure metal and offers more resistance. Moreover, at high temperatures, the alloys do not melt. Hence, the coils of heating appliances such as electric toasters and electric irons are made of an alloy rather than a pure metal.
(a) The Resistivity of mercury is more than that of iron. So mercury offer more resistance than that iron whereas iron have lower resistivity and offer less resistance therefore iron is a better conductor than mercury
Resistivity of iron = 10.0 x 10-8 
Resistivity of mercury = 94.0 × 10−8
(b) Silver is the best conductor of electricity because of lowest resistivity and offers lower resistance as shown according to the above table.
In a given circuit diagram here shows three resistors of resistances 5 Ω, 8 Ω and 12 Ω respectively that are connected in series and a battery of potential 6 V.Three cells of potential 2 V, each connected in series, is equivalent to a battery of potential 2 V + 2 V + 2 V = 6V.
Current in a circuit is measured through an ammeter that should be connected in the circuit in series with the resistors. To measure the potential difference across the 12 Ω resistor, a voltmeter should be connected parallel to this resistor, as shown in the following figure.
The resistances are connected in series.
Ohm’s law can be used to obtain the readings of ammeter and voltmeter. According to it
V = IR,
Here ,
V Potential difference= 6 V
Current flowing through the circuit/resistors = I
Resistance of the circuit, R = 5+8+12 = 25 Ω
Potential difference across 12 Ω resistor = VI
Current flowing through the 12 Ω resistor, I = 0.24 A
Therefore, using Ohm’s law, we obtain
VI = IR = 0.24x12 = 2.88 V
So the reading of the ammeter will be 0.24 A.
The reading of the voltmeter will be 2.88 V.
(a) When 1 Ω and 106 Ω are connected in parallel:
Let R be the equivalent resistance.
Therefore, equivalent resistance 1 Ω
(b) When 1 Ω, and are connected in parallel:
Let R be the equivalent resistance.
Therefore, equivalent resistance = 0.999 Ω
Resistance of electric lamp, R1 = 100 Ω
Resistance of toaster, R2 = 50 Ω
Resistance of water filter, R3 = 500 Ω
Voltage of the source, V = 220 V
These are connected in parallel, as shown in the following figure.
Let R be the equivalent resistance of the circuit.
According to Ohm’s law,
V = IR
Where,
Current flowing through the circuit = I
7.04 A of current is drawn by all the three given appliances.
Therefore, current drawn by an electric iron connected to the same source of potential 220 V = 7.04 A
Let R' be the resistance of the electric iron. According to Ohm’s law,
Therefore, the resistance of the electric iron is 31.25 Ω and the current flowing through it is 7.04 A.
There is no division of voltage among the appliances when connected in parallel. The potential difference across each appliance is equal to the supplied voltage.
The total effective resistance of the circuit can be reduced by connecting electrical appliances in parallel.
There are three resistors of resistances 2 Ω, 3 Ω, and 6 Ω respectively.
(a) The following circuit diagram shows the connection of the three resistors.
Here, 6 Ω and 3 Ω resistors are connected in parallel.
Therefore, their equivalent resistance will be given by
This equivalent resistor of resistance 2 Ω is connected to a 2 Ω resistor in series.
Therefore, equivalent resistance of the circuit = 2 Ω + 2 Ω = 4 Ω
Hence, the total resistance of the circuit is 4 Ω.
(b) The following circuit diagram shows the connection of the three resistors.
All the resistors are connected in series. Therefore, their equivalent resistance will be given as
Therefore, the total resistance of the circuit is 1 Ω.
There are four coils of resistances 4 Ω, 8 Ω, 12 Ω, and 24 Ω respectively.
(a) If these coils are connected in series, then the equivalent resistance will be the highest, given by the sum 4 + 8 + 12 + 24 = 48 Ω
(b) If these coils are connected in parallel, then the equivalent resistance will be the lowest, given by
Therefore, 2 Ω is the lowest total resistance.
The heating element of an electric heater is a resistor. The amount of heat produced by it is proportional to its resistance. The resistance of the element of an electric heater is very high. As current flows through the heating element, it becomes too hot and glows red. On the other hand, the resistance of the cord is low. It does not become red when current flows through it.
The amount of heat (H) produced is given by the Joule’s law of heating as
H = VIt
Where,
Voltage, V = 50 V
Time, t = 1 h = 1 × 60 × 60 s
Amount of current,
Therefore, the heat generated is 4.8 x 106 J.
The amount of heat (H) produced is given by the joule’s law of heating as
H = VIt
Where,
Current, I = 5 A
Time, t = 30 s
Voltage, V = Current × Resistance = 5 × 20 = 100 V
H = 100 x 5 x 30 = 1.5x104 J
Therefore, the amount of heat developed in the electric iron is 1.5x104 J.
The rate of consumption of electric energy in an electric appliance is called electric power. Hence, the rate at which energy is delivered by a current is the power of the appliance.
Power (P) is given by the expression,
P = VI
Where,
Voltage, V = 220 V
Current, I = 5 A
P = 220 x 5 = 1100 W
Energy consumed by the motor = Pt
Where,
Time, t = 2 h = 2 × 60 × 60 = 7200 s
P = 1100 × 7200 = 7.92 × 106 J
Therefore, power of the motor = 1100 W
Energy consumed by the motor = 7.92 × 106 J
(d) Resistance of a piece of wire is proportional to its length. A piece of wire has a resistance R. The wire is cut into five equal parts.
Therefore, resistance of each part = 
All the five parts are connected in parallel. Hence, equivalent resistance (R’) is given as
Therefore, the ratio 
is 25.
(b) Electrical power is given by the expression, P = VI … (i)
According to Ohm’s law, V = IR … (ii)
Where,
V = Potential difference
I = Current
R = Resistance
P = VI
From equation (i), it can be written
P = (IR) × I
P = I2R
From equation (ii), it can be written
Power P cannot be expressed as IR2.
(d) Energy consumed by an appliance is given by the expression,
Where,
Power rating, P = 100 W
Voltage, V = 220 V
The resistance of the bulb remains constant if the supply voltage is reduced to 110 V. If the bulb is operated on 110 V, then the energy consumed by it is given by the expression for power as
Therefore, the power consumed will be 25 W.
(c) The Joule heating is given by, H = i2Rt
Let, R be the resistance of the two wires.
The equivalent resistance of the series connection is RS = R + R = 2R
If V is the applied potential difference, then it is the voltage across the equivalent resistance.
The heat dissipated in time t is,
The equivalent resistance of the parallel connection is
V is the applied potential difference across this RP.
The heat dissipated in time t is,
So, the ratio of heat produced is,
Note: H 
also H i2 and H t. In this question, t is same for both the circuit. But the current through the equivalent resistance of both the circuit is different. We could have solved the question directly using H R if in case the current was also same. As we know the voltage and resistance of the circuits, we have calculated i in terms of voltage and resistance and used in the equation H = i2Rt to find the ratio.
To measure the potential difference between two points, a voltmeter should be connected in parallel to the points.
Resistance (R) of a copper wire of length l and cross-section A is given by the expression,
Where,
Resistivity of copper, p = 1.6 x 10-8 Ω m
Area of cross-section of the wire,
Diameter = 0.5 mm = 0.0005 m
Resistance, R = 10 Ω
Hence, length of the wire,
If the diameter of the wire is doubled, new diameter = 2x0.5 = 1 mm = 0.001 m
Therefore, resistance R'
Therefore, the length of the wire is 122.7 m and the new resistance is 2.5 Ω
The plot between voltage and current is called IV characteristic. The voltage is plotted on x-axis and current is plotted on y-axis. The values of the current for different values of the voltage are shown in the given table.
The IV characteristic of the given resistor is plotted in the following figure.
The slope of the line gives the value of resistance (R) as,
Therefore, the resistance of the resistor is 3.4 Ω.
Resistance (R) of a resistor is given by Ohm’s law as,
Where,
Potential difference, V = 12 V
Current in the circuit, I = 2.5 mA = 2.5x10-3 A
Therefore, the resistance of the resistor is 4.8 kΩ.
There is no current division occurring in a series circuit. Current flow through the component is the same, given by Ohm’s law as
Where,
R is the equivalent resistance of resistances 0.2 Ω, 0.3 Ω, 0.4 Ω, 0.5 Ω and 12 Ω. These are connected in series. Hence, the sum of the resistances will give the value of R.
R = 0.2 + 0.3 + 0.4 + 0.5 + 12 = 13.4 Ω
Potential difference, V = 9 V
Therefore, the current that would flow through the 12 Ω resistor is 0.671 A.
For x number of resistors of resistance 176 Ω, the equivalent resistance of the resistors connected in parallel is given by Ohm’s law as
Where,
Supply voltage, V = 220 V
Current, I = 5 A
Equivalent resistance of the combination = R, given as
From Ohm’s law,
Therefore, four resistors of 176 Ω are required to draw the given amount of current.
If we connect the resistors in series, then the equivalent resistance will be the sum of the resistors, i.e., 6 Ω + 6 Ω + 6 Ω = 18 Ω, which is not desired. If we connect the resistors in parallel, then the equivalent resistance will be
which is also not desired. Hence, we should either connect the two resistors in series or parallel.
(i) Two resistors in parallel
Two 6 Ω resistors are connected in parallel. Their equivalent resistance will be
The third 6 Ω resistor is in series with 3 Ω. Hence, the equivalent resistance of the circuit is 6 Ω + 3 Ω = 9 Ω.
(ii) Two resistors in series
Two 6 Ω resistors are in series. Their equivalent resistance will be the sum 6 + 6 = 12 Ω
The third 6 Ω resistor is in parallel with 12 Ω. Hence, equivalent resistance will be
Therefore, the total resistance is 4 Ω.
Resistance R1 of the bulb is given by the expression,
Where,
Supply voltage, V = 220 V
Maximum allowable current, I = 5 A
Rating of an electric bulb P1 = 10 W
According to Ohm’s law,
V = I R
Where,
R is the total resistance of the circuit for x number of electric bulbs
Resistance of each electric bulb, R1 = 4840 Ω
Therefore, 110 electric bulbs are connected in parallel.
Supply voltage, V = 220 V
Resistance of one coil, R = 24 Ω
(i) Coils are used separately
According to Ohm’s law,
V = I1R1
Where,
I1 is the current flowing through the coil
Therefore, 9.16 A current will flow through the coil when used separately.
(ii) Coils are connected in series
Total resistance, R2 = 24 Ω + 24 Ω = 48 Ω
According to Ohm’s law,
V = I2R2
Where,
I2 is the current flowing through the series circuit
Therefore, 4.58 A current will flow through the circuit when the coils are connected in series.
(iii) Coils are connected in parallel
Total resistance, R3 is given as
According to Ohm’s law,
V = I3R3
Where,
I3 is the current flowing through the circuit
Therefore, 18.33 A current will flow through the circuit when coils are connected in parallel.
(i) Potential difference, V = 6 V
1 Ω and 2 Ω resistors are connected in series. Therefore, equivalent resistance of the circuit, R = 1 + 2 = 3 Ω
According to Ohm’s law,
V = IR
Where,
I is the current through the circuit
This current will flow through each component of the circuit because there is no division of current in series circuits. Hence, current flowing through the 2 Ω resistor is 2 A. Power is given by the expression,
P = (I)2R = (2)2 x 2 = 8 W
(ii) Potential difference, V = 4 V
12 Ω and 2 Ω resistors are connected in parallel. The voltage across each component of a parallel circuit remains the same. Hence, the voltage across 2 Ω resistor will be 4 V.
Power consumed by 2 Ω resistor is given by
Therefore, the power used by 2 Ω resistor is 8 W.
Both the bulbs are connected in parallel. Therefore, potential difference across each of them will be 220 V, because no division of voltage occurs in a parallel circuit.
Current drawn by the bulb of rating 100 W is given by,
Similarly, current drawn by the bulb of rating 60 W is given by,
Energy consumed by an electrical appliance is given by the expression,
H = Pt
Where,
Power of the appliance = P
Time = t
Energy consumed by a TV set of power 250 W in 1 h = 250 × 3600 = 9 × 105 J
Energy consumed by a toaster of power 1200 W in 10 minutes = 1200 × 600
= 7.2× 105 J
Therefore, the energy consumed by a 250 W TV set in 1 h is more than the energy consumed by a toaster of power 1200 W in 10 minutes.
Rate of heat produced by a device is given by the expression for power as
P = I2R
Where,
Resistance of the electric heater, R = 8 Ω
Current drawn, I = 15 A
P = (15)2 x 8 = 1800 J/s
Therefore, heat is produced by the heater at the rate of 1800 J/s.
(a) The melting point and resistivity of tungsten are very high. It does not burn readily at a high temperature. The electric lamps glow at very high temperatures. Hence, tungsten is mainly used as heating element of electric bulbs.
(b) The conductors of electric heating devices such as bread toasters and electric irons are made of alloy because resistivity of an alloy is more than that of metals. It produces large amount of heat.
(c) There is voltage division in series circuits. Each component of a series circuit receives a small voltage for a large supply voltage. As a result, the amount of current decreases and the device becomes hot. Hence, series arrangement is not used in domestic circuits.
(d) Resistance (R) of a wire is inversely proportional to its area of cross-section (A), i.e.,
(e) Copper and aluminium wires have low resistivity. They are good conductors of electricity. Hence, they are usually employed for electricity transmission.
